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Work done by force (Vector notation)

by Gold3nlily
Tags: force, notation, vector, work
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Gold3nlily
#1
Apr3-11, 11:22 AM
P: 42
1. The problem statement, all variables and given/known data

How much work is done by a force= (2x)i N + (3)j N with x in meters, that moves a particle from a position (initial) = (2m)i + (3m)j to a position (final) = (-4m)i + (-3m)j?

Answer:
- 6 J

2. Relevant equations

Work = integral of Force

3. The attempt at a solution

W = [x^2] i + [3x] j
W = [4^2 - 2^2] i + [3(-3) - 3(3)] j
W = 12 i -18 j
W magnitude = 21.6J

I realize I am doing this wrong becasue the Magnitude will always be positive and the answer (from back of book) is negative.

I tried this in a similar way where I took the difference between the two position vector and then placed it in the integral
rd = (-6) i + (-6) j
Got W = 36 i + 18 j
still wrong.

Help, pretty please?
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Rasalhague
#2
Apr3-11, 12:20 PM
P: 1,402
To save typing all the i's and j's, I'll use the notation (2,3) = 2i + 3j. I'll also cut corners by leaving out the units.

Assuming the movement is in a straight line, we can parameterize the particle's path from the initial point (2,3) to the final point (-4,-3) by an equation such as

[tex]\mathbf{r}(t) = (x,y) = (2,3) - t(6,6), \enspace 0 \leq t \leq 1.[/tex]

-(6,6) is the displacement vector from the initial point to the final point, that is, the difference between them.

Then [itex]x=2-6t[/itex], and

[tex]d\mathbf{r} = (dx,dy) = -(6,6) dt.[/tex]

So

[tex]\int_{\mathbf{r}(0)}^{\mathbf{r}(1)} \mathbf{F}(\mathbf{r}) \cdot d\mathbf{r}[/tex]

[tex]=\int_{(2,3)}^{(-4,-3)}(2x,3) \cdot (dx,dy) = \int_{0}^{1}(42-72t) \; dt = -6[/tex]
Gold3nlily
#3
Apr3-11, 01:16 PM
P: 42
Quote Quote by Rasalhague View Post

[tex]=\int_{(2,3)}^{(-4,-3)}(2x,3) \cdot (dx,dy) = \int_{0}^{1}(42-72t) \; dt = -6[/tex]
Thank you Rasalhague. But how did you get from (2x,3) to (42-72t)? I don't understand.

Rasalhague
#4
Apr3-11, 02:21 PM
P: 1,402
Work done by force (Vector notation)

Oopsh, sorry, I missed out a minus in front of the integral on the right. Are the following steps clear?

[tex]x = 2 - 6t[/tex]

[tex](2x,3) = (2(2-6t),3) = (4-12t,3)[/tex]

[tex](4-12t,3) \cdot (-6,-6) = -6(4-12t) - 18 = -42 + 72t.[/itex]

-(6,6) comes from differentiating [itex]\mathbf{r}[/itex].
Gold3nlily
#5
Apr9-11, 05:07 PM
P: 42
Quote Quote by Rasalhague View Post
Oopsh, sorry, I missed out a minus in front of the integral on the right. Are the following steps clear?

[tex]x = 2 - 6t[/tex]

[tex](2x,3) = (2(2-6t),3) = (4-12t,3)[/tex]

[tex](4-12t,3) \cdot (-6,-6) = -6(4-12t) - 18 = -42 + 72t.[/itex]

-(6,6) comes from differentiating [itex]\mathbf{r}[/itex].
Hi Rasalhague. Thank you for helping me.

My test is this Wednesday, but I have lots of other tests so sorry about the delay.

A couple things:

1) where did time come from? The question never mentioned it- only position.

2) I understand this:
r = (-6) i + (-6) j
r = -(6,6)

3) where does this come from: x=2-6t (I think this is an extension of being confused about the time)?

Why is integral not (x^2)i + (3x) j ???

Thank you again for your help and I am sorry if I am difficult.
Gold3nlily
#6
Apr10-11, 09:24 AM
P: 42
The reason time confuses me is that isn't POWER change in work over time? I thought the difference between Force and power is whether your taking the derivative of work with respect to time or with respect to position....?
Gold3nlily
#7
Apr10-11, 01:10 PM
P: 42
Okay, I figured out what I did wrong. I had this part right:

W = [x^2] i + [3x] j
W = [4^2 - 2^2] i + [3(-3) - 3(3)] j
W = 12 i -18 j

What I needed to do was subtract 18 from 12
W = 12 - 18 = -6 (answer!)

I wasn't doing this becasue I thought they were sepearte becasue of the separate components. Wow, I can't believe I was so close for so long!

I still don't completely understand why you just add i and j together to find work instead of finding the magnitude... but it works for many similar book problems so I know this is the right way to do it.

Thanks for the help. :)


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